ML Aggarwal Solutions for Class 9 Maths Chapter 7 – Quadratic Equations are provided here to help students understand all the concepts clearly and develop a strong command over the subject. This chapter mainly deals with problems based on quadratic equations. Sometimes, students face difficulty in understanding the concepts during class hours, for those students, the experts at BYJU’S have designed the solutions based on the students’ grasping abilities. Students can solve both chapter-wise and exercise-wise problems to increase their confidence level before appearing for the board exam. To boost interest among students in this chapter, ML Aggarwal Solutions for Class 9 Chapter 7 quadratic equations PDF links are given here for easy access. Students can download the PDF easily and can use it for future reference as well.
Chapter 7 – Quadratic Equations contains one exercise and ML Aggarwal Class 9 Solutions present on this page provide solutions to questions related to each topic discussed in this chapter.
ML Aggarwal Solutions for Class 9 Maths Chapter 7 – Quadratic Equations
Access answers to ML Aggarwal Solutions for Class 9 Maths Chapter 7 – Quadratic Equations
EXERCISE 7
Solve the following (1 to 12) equations:
1. (i) x² – 11x + 30 = 0
(ii) 4x² – 25 = 0
Solution:
(i) x² – 11x + 30 = 0
Let us simplify the given equation,
By factorizing, we get
x2 – 5x – 6x + 30 = 0
x(x – 5) – 6 (x – 5) = 0
(x – 5) (x – 6) = 0
So,
(x – 5) = 0 or (x – 6) = 0
x = 5 or x = 6
∴ Value of x = 5, 6
(ii) 4x² – 25 = 0
Let us simplify the given equation,
4x² = 25
x2 = 25/4
x = ± √(25/4)
= ±5/2
∴ Value of x = +5/2, -5/2
2. (i) 2x² – 5x = 0
(ii) x² – 2x = 48
Solution:
(i) 2x² – 5x = 0
Let us simplify the given equation,
x(2x – 5) = 0
so,
x = 0 or 2x – 5 = 0
x = 0 or 2x = 5
x = 0 or x = 5/2
∴ Value of x = 0, 5/2
(ii) x² – 2x = 48
Let us simplify the given equation,
By factorizing, we get
x2 – 2x – 48 = 0
x2 – 8x+ 6x – 48 = 0
x(x – 8) + 6 (x – 8) = 0
(x – 8) (x + 6) = 0
So,
(x – 8) = 0 or (x + 6) = 0
x = 8 or x = -6
∴ Value of x = 8, -6
3. (i) 6 + x = x²
(ii) 2x² + 3x + 1= 0
Solution:
(i) 6 + x = x²
Let us simplify the given equation,
6 + x – x2 = 0
x2 – x – 6 = 0
By factorizing, we get
x2 – 3x + 2x – 6 = 0
x(x – 3) + 2 (x – 3) = 0
(x – 3) (x + 2) = 0
So,
(x – 3) = 0 or (x + 2) = 0
x = 3 or x = -2
∴ Value of x = 3, -2
(ii) 2x² + 3x + 1= 0
Let us simplify the given equation,
By factorizing, we get
2x2 – 2x – x + 1 = 0
2x(x – 1) – 1 (x – 1) = 0
(x – 1) (2x – 1) = 0
So,
(x – 1) = 0 or (2x – 1) = 0
x = 1 or 2x = 1
x = 1 or x = ½
∴ Value of x = 1, ½
4. (i) 3x² = 2x + 8
(ii) 4x² + 15 = 16x
Solution:
(i) 3x² = 2x + 8
Let us simplify the given equation,
3x2 – 2x – 8 = 0
By factorizing, we get
3x2 – 6x + 4x – 8 = 0
3x(x – 2) + 4 (x – 2) = 0
(x – 2) (3x + 4) = 0
So,
(x – 2) = 0 or (3x + 4) = 0
x = 2 or 3x = -4
x = 2 or x = -4/3
∴ Value of x = 2 or -4/3
(ii) 4x² + 15 = 16x
Let us simplify the given equation,
4x2 – 16x + 15 = 0
By factorizing, we get
4x2 – 6x – 10x + 15 = 0
2x(2x – 3) – 5 (2x – 3) = 0
(2x – 3) (2x – 5) = 0
So,
(2x – 3) = 0 or (2x – 5) = 0
2x = 3 or 2x = 5
x = 3/2 or x = 5/2
∴ Value of x = 3/2 or 5/2
5. (i) x (2x + 5) = 25
(ii) (x + 3) (x – 3) = 40
Solution:
(i) x (2x + 5) = 25
Let us simplify the given equation,
2x2 + 5x – 25 = 0
By factorizing, we get
2x2 + 10x – 5x – 25 = 0
2x(x + 5) – 5 (x + 5) = 0
(x + 5) (2x – 5) = 0
So,
(x + 5) = 0 or (2x – 5) = 0
x = -5 or 2x = 5
x = -5 or x = 5/2
∴ Value of x = -5, 5/2
(ii) (x + 3) (x – 3) = 40
Let us simplify the given equation,
x2 – 3x + 3x – 9 = 40
x2 – 9 – 40 = 0
x2 – 49 = 0
x2 = 49
x = √49
= ± 7
∴ Value of x = 7, -7
6. (i) (2x + 3) (x – 4) = 6
(ii) (3x + 1) (2x + 3) = 3
Solution:
(i) (2x + 3) (x – 4) = 6
Let us simplify the given equation,
2x2 – 8x + 3x – 12 – 6 = 0
2x2 – 5x – 18 = 0
By factorizing, we get
2x2 – 9x + 4x – 18 = 0
x (2x – 9) + 2 (2x – 9) = 0
(2x – 9) (x + 2) = 0
So,
(2x – 9) = 0 or (x + 2) = 0
2x = 9 or x = -2
x = 9/2 or x = -2
∴ Value of x = 9/2, -2
(ii) (3x + 1) (2x + 3) = 3
Let us simplify the given equation,
6x2 + 9x + 2x + 3 – 3 = 0
6x2 + 11x = 0
x(6x + 11) = 0
So,
x = 0 or 6x + 11 = 0
x = 0 or 6x = -11
x = 0 or x = -11/6
∴ Value of x = 0, -11/6
7. (i) 4x² + 4x + 1 = 0
(ii) (x – 4)² + 5² = 132
Solution:
(i) 4x² + 4x + 1 = 0
Let us simplify the given equation,
By factorizing, we get
4x2 + 2x + 2x + 1 = 0
2x(2x + 1) + 1 (2x + 1) = 0
(2x + 1) (2x + 1) = 0
So,
(2x + 1) = 0 or (2x + 1) = 0
2x = -1 or 2x = -1
x = -1/2 or x = -1/2
∴ Value of x = -1/2, -1/2
(ii) (x – 4)² + 5² = 132
Let us simplify the given equation,
x2 + 16 – 2(x) (4) + 25 – 169 = 0
x2 – 8x -128 = 0
By factorizing, we get
x2 – 16x + 8x – 128 = 0
x(x – 16) + 8 (x – 16) = 0
(x – 16) (x + 8) = 0
So,
(x – 16) = 0 or (x + 8) = 0
x = 16 or x = -8
∴ Value of x = 16, -8
8. (i) 21x2 = 4 (2x + 1)
(ii) 2/3x2 – 1/3x – 1 = 0
Solution:
(i) 21x2 = 4 (2x + 1)
Let us simplify the given equation,
21x2 = 8x + 4
21x2 – 8x – 4 = 0
By factorizing, we get
21x2 – 14x + 6x – 4 = 0
7x(3x – 2) + 2(3x – 2) = 0
(3x – 2) (7x + 2) = 0
So,
(3x – 2) = 0 or (7x + 2) = 0
3x = 2 or 7x = -2
x = 2/3 or x = -2/7
∴ Value of x = 2/3 or -2/7
(ii) 2/3x2 – 1/3x – 1 = 0
Let us simplify the given equation,
By taking 3 as LCM and cross multiplying
2x2 – x – 3 = 0
By factorizing, we get
2x2 – 3x + 2x – 3 = 0
x(2x – 3) + 1 (2x – 3) = 0
(2x – 3) (x + 1) = 0
So,
(2x – 3) = 0 or (x + 1) = 0
2x = 3 or x = -1
x = 3/2 or x = -1
∴ Value of x = 3/2, -1
9. (i) 6x + 29 = 5/x
(ii) x + 1/x = 2 ½
Solution:
(i) 6x + 29 = 5/x
Let us simplify the given equation,
By cross multiplying, we get
6x2 + 29x – 5 = 0
By factorizing, we get
6x2 + 30x – x – 5 = 0
6x (x + 5) -1 (x + 5) = 0
(x + 5) (6x – 1) = 0
So,
(x + 5) = 0 or (6x – 1) = 0
x = -5 or 6x = 1
x = -5 or x = 1/6
∴ Value of x = -5, 1/6
(ii) x + 1/x = 2 ½
x + 1/x = 5/2
Let us simplify the given equation,
By taking LCM
x2 + 1 = 5x/2
By cross multiplying,
2x2 + 2 – 5x = 0
2x2 – 5x + 2 = 0
By factorizing, we get
2x2 – x – 4x + 2 = 0
x(2x – 1) – 2 (2x – 1) = 0
(2x – 1) (x – 2) = 0
So,
(2x – 1) = 0 or (x – 2) = 0
2x = 1 or x = 2
x = ½ or x = 2
∴ Value of x = ½, 2
10. (i) 3x – 8/x = 2
(ii) x/3 + 9/x = 4
Solution:
(i) 3x – 8/x = 2
Let us simplify the given equation,
By taking LCM and cross multiplying,
3x2 – 8 = 2x
3x2 – 2x – 8 = 0
By factorizing, we get
3x2 – 6x + 4x – 8 = 0
3x(x – 2) + 4 (x – 2) = 0
(x – 2) (3x + 4) = 0
So,
(x – 2) = 0 or (3x + 4) = 0
x = 2 or 3x = -4
x = 2 or x = -4/3
∴ Value of x = 2, -4/3
(ii) x/3 + 9/x = 4
Let us simplify the given equation,
By taking 3x as LCM and cross multiplying
x2 + 27 = 12x
x2 – 12x + 27 = 0
By factorizing, we get
x2 – 3x – 9x + 27 = 0
x (x – 3) – 9 (x – 3) = 0
(x – 3) (x – 9) = 0
So,
(x – 3) = 0 or (x – 9) = 0
x = 3 or x = 9
∴ Value of x = 3, 9
11. (i) (x – 1)/(x + 1) = (2x – 5)/(3x – 7)
(ii) 1/(x + 2) + 1/x = ¾
Solution:
(i) (x – 1)/(x + 1) = (2x – 5)/(3x – 7)
Let us simplify the given equation,
By cross multiplying,
(x – 1) (3x – 7) = (2x – 5) (x + 1)
3x2 – 7x – 3x + 7 = 2x2 + 2x – 5x – 5
3x2 – 10x + 7 – 2x2 +3x + 5 = 0
x2 – 7x + 12 = 0
By factorizing, we get
x2 – 4x – 3x + 12 = 0
x (x – 4) – 3 (x – 4) = 0
(x – 4) (x – 3) = 0
So,
(x – 4) = 0 or (x – 3) = 0
x = 4 or x = 3
∴ Value of x = 4, 3
(ii) 1/(x + 2) + 1/x = ¾
Let us simplify the given equation,
By taking x(x + 2) as LCM
(x+x+2)/x(x + 2) = ¾
By cross multiplying,
4(2x + 2) = 3x(x + 2)
8x + 8= 3x2 + 6x
3x2 + 6x – 8x – 8 = 0
3x2 – 2x – 8 = 0
By factorizing, we get
3x2 – 6x + 4x – 8 = 0
3x(x – 2) + 4 (x – 2) = 0
(x – 2) (3x + 4) = 0
So,
(x – 2) = 0 or (3x + 4) = 0
x = 2 or 3x = -4
x = 2 or x = -4/3
∴ Value of x = 2, -4/3
12. (i) 8/(x + 3) – 3/(2 – x) = 2
(ii) x/(x + 1) + (x + 1)/x = 2 1/6
Solution:
(i) 8/(x + 3) – 3/(2 – x) = 2
Let us simplify the given equation,
By taking (x+3)(2-x) as LCM
[8(2-x) – 3(x+3)] / (x+3) (2-x) = 2 [16 – 8x – 3x – 9] / [2x – x2 + 6 – 3x] = 2 [-11x + 7] = 2(-x2 – x + 6)7 – 11x = -2x2 – 2x + 12
2x2 + 2x – 11 x – 12 + 7 = 0
2x2 – 9x – 5 = 0
By factorizing, we get
2x2 – 10x + x – 5 = 0
2x (x – 5) + 1 (x – 5) = 0
(x – 5) (2x + 1) = 0
So,
(x – 5) = 0 or (2x + 1) = 0
x = 5 or 2x= -1
x = 5 or x = -1/2
∴ Value of x = 5, -1/2
(ii) x/(x + 1) + (x + 1)/x = 2 1/6
x/(x + 1) + (x + 1)/x = 13/6
Let us simplify the given equation,
By taking x(x+1) as LCM
[x(x) + (x+1) (x+1)] / x(x + 1) = 13/66[x2 + x2 + x + x + 1] = 13x(x + 1)
6[2x2 + 2x + 1] = 13x2 + 13x
12x2 + 12x + 6 – 13x2 – 13x = 0
-x2 – x + 6 = 0
x2 + x – 6 = 0
By factorizing, we get
x2 + 3x – 2x – 6 = 0
x (x + 3) – 2 (x + 3) = 0
(x + 3) (x – 2) = 0
So,
(x + 3) = 0 or (x – 2) = 0
x = -3 or x = 2
∴ Value of x = -3, 2
Chapter Test
Solve the following (1 to 3) equations:
1. (i) x (2x + 5) = 3
(ii) 3x2 – 4x – 4 = 0.
Solution:
(i) x (2x + 5) = 3
We can write it as
2x2 + 5x – 3 = 0
By further calculation
2x2 + 6x – x – 3 = 0
By taking out the common terms
2x (x + 3) – 1 (x + 3) = 0
So we get
(x + 3) (2x – 1) = 0
Here
x + 3 = 0 then x = – 3
2x – 1 = 0 then 2x = 1 where x = ½
Therefore, x = – 3, ½.
(ii) 3x2 – 4x – 4 = 0
We can write it as
3x2 – 6x + 2x – 4 = 0
By taking out the common terms
3x (x – 2) + 2 (x – 2) = 0
So we get
(x – 2) (3x + 2) = 0
Here
x – 2 = 0 then x = 2
3x + 2 = 0 then 3x = – 2 where x = – 2/3
Therefore, x = 2, – 2/3.
2. (i) 4x2 – 2x + ¼ = 0
(ii) 2x2 + 7x + 6 = 0.
Solution:
(i) 4x2 – 2x + ¼ = 0
Multiply the equation by 4
16x2 – 8x + 1 = 0
We can write it as
16x2 – 4x – 4x + 1 = 0
Taking out the common terms
4x (4x – 1) – 1 (4x – 1) = 0
So we get
(4x – 1) (4x – 1) = 0
(4x – 1)2 = 0
Here
4x – 1 = 0
4x = 1
By division
x = ¼, ¼
(ii) 2x2 + 7x + 6 = 0
We can write it as
2x2 + 4x + 3x + 6 = 0
By further calculation
2x (x + 2) + 3 (x + 2) = 0
So we get
(x + 2) (2x + 3) = 0
Here
x + 2 = 0 then x = – 2
2x + 3 = 0 then 2x = – 3 where x = – 3/2
x = – 2, – 3/2
3. (i) (x – 1)/ (x – 2) + (x – 3)/ (x – 4) = 3 1/3
(ii) 6/x – 2/(x – 1) = 1/(x – 2).
Solution:
(i) (x – 1)/ (x – 2) + (x – 3)/ (x – 4) = 3 1/3
By taking LCM
[(x – 1) (x – 4) + (x – 2) (x – 3)]/ (x – 2) (x – 4) = 10/3By further calculation
(x2 – 5x + 4 + x2 – 5x + 6)/ (x2 – 6x + 8) = 10/3
So we get
(2x2 – 10x + 10)/ (x2 – 6x + 8) = 10/3
By cross multiplication
10x2 – 60x + 80 = 6x2 – 30x + 30
By further simplification
10x2 – 60x + 80 – 6x2 + 30x – 30 = 0
So we get
4x2 – 30x + 50 = 0
Dividing by 2
2x2 – 15x + 25 = 0
It can be written as
2x2 – 10x – 5x + 25 = 0
Taking out the common terms
2x (x – 5) – 5 (x – 5) = 0
(x – 5) (2x – 5) = 0
Here
x – 5 = 0 then x = 5
2x – 5 = 0 then 2x = 5 where x = 5/2
Therefore, x = 5, 5/2.
(ii) 6/x – 2/(x – 1) = 1/(x – 2)
Taking LCM
(6x – 6 – 2x)/ x (x – 1) = 1/ (x – 2)
By further calculation
(4x – 6)/ (x2 – x) = 1/(x – 2)
By cross multiplication
4x2 – 8x – 6x + 12 = x2 – x
So we get
4x2 – 14x + 12 – x2 + x = 0
3x2 – 13x + 12 = 0
3x2 – 4x – 9x + 12 = 0
Taking out the common terms
x (3x – 4) – 3 (3x – 4) = 0
(3x – 4) (x – 3) = 0
Here
3x – 4 = 0 then 3x = 4 where x = 4/3
x – 3 = 0 then x = 3
Therefore, x = 3, 4/3.
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